First, you need to start your sum at $n=1$, not $n=0$.
To address your first question:
There is no contradiction. If a series converges uniformly on a set $O$, then it converges uniformly on any subset of $O$; but, the series does not necessarily converge uniformly on a set $U\supset O$.
You have uniform convergence on a set $(a,\infty)$, $a>0$ and non-uniform convergence on a larger set $(0,\infty)$, which is ok...
To, hopefully, address your second question:
In your example, you have uniform convergence on sets of the form $(a,\infty)$, $a>0$.
You have pointwise convergence on the set $(0,\infty)$. Moreover, since the series diverges at $x=0$, this is the largest interval, unbounded on the right, for which you have pointwise convergence. This is then
the natural candidate for an interval over which you could have non-uniform convergence.
Moreover, things would have to go awry near $x=0$.
But is this indeed the case?
Let's look at the graphs of the first few partial sums $s_n=\sum\limits_{k=1}^n {1\over k(1+kx^2). }$

We are led to suspect that the series does not converge uniformly on $(0,\infty)$. (look at how the difference between the $s_k$ grows larger as $x\rightarrow0^+$).
In fact, given $N$, we may choose $M>N$ so that $\sum\limits_{n=N}^M{1\over n}\ge 1$. Then
$$
\lim_{x\rightarrow0^+}\sum_{n=N}^M {1\over n(1+nx^2)}
=\lim_{x\rightarrow0^+}\sum_{n=N}^M {1\over n}\ge1.
$$
So the series $\sum\limits_{n=1}^\infty {1\over n(1+nx^2)}$ is not uniformly Cauchy on $(0,\infty)$ and thus not uniformly convergent on $(0,\infty)$.
Since $x=0$ is the "bad point" as far as uniform convergence is concerned, the series does not converge uniformly on any interval of the form $(0,a)$.